**james@umcp-cs.UUCP** (11/03/83)

Try this: What topological space has the fundamental group {e , 1} ? By {e , 1}, I mean the (only) two-element group, otherwise known as the integers mod 2, under addition. Fundamental Group: The fundamental group of a pathwise-connected topological space is the group of equivalence classes of loops in the space. A loop is a continuous map of the interval [0,1] into the space (where 0 and 1 are mapped to the same point, to close the loop). Two loops are equivalent iff they can be continuously deformed into each other (they are homotopic.) The group operation is by composition of loops (i.e. you map [0,.5] onto the first loop, and [.5,1] onto the second loop. --Jim O'Toole

**ech@pyuxnn.UUCP** (11/11/83)

Since no one else seems to responded to the request for *a* ( not *the* ) topological space having Z/2Z ( the integers modulo 2) as fundamental group, I will do so. The simplest example is the real projective plane,RP(2), which has figured in previous articles in net.math. For the purposes of this discussion, we may think of RP(2) as the space which results when we identify antipodal points on the 2-sphere,S. The map p:S --> RP(2) which sends any point in S to the corresponding equivalence class in RP(2) is a *covering map*,i.e. each point in RP(2) has a neighborhood U such that the inverse image of U under p is a disjoint union of open sets in S, each of which is mapped homeomorphically by the restriction of p. The study of such maps forms an interesting chapter in elementary algebraic topology; we summarize some of the results below. Choose a basepoint x0 in RP(2) & then select a basepoint e0 in S so that p(e0)=x0. Given a loop f:[0;1] --> RP(2) based at x0 (i.e. f(0)=f(1)=x0), it is easy to show that f possesses a *lifting* to a map f':[0;1] --> S such that f'(0)=e0 and pf'=f and, indeed, this map f' is *unique*. It is somewhat more difficult to prove that any homotopy of loops can be lifted, but it's true; if F:[0;1]x[0;1] --> RP(2) is a homotopy of loops based at x0 ( so that F(0,t)=F(1,t)=x0 for all t ), then there is a unique map F':[0;1]x[0;1] --> S such that F'(0,0)=e0 and pF'=F. Geometrically, these facts say that any loop in RP(2) based at x0 is covered by a path in S which connects e0 to some point in the *fiber* over x0 ( possibly different from e0 ) and that this second endpoint of the lifted loop depends only on the homotopy class of the loop with which we began. This procedure thus defines a function with domain the fundamental group of (RP(2),x0) and range the fiber over x0. Because S is path-connected, this function is *onto* the fiber over x0; given a point e in the fiber, any path from e0 to e in S projects to a loop at p(e)=p(e0)=x0 whose homotopy class will be sent by our function to e. Because S is *simply-connected*, this function is also *one-to-one*; this quickly reduces to showing that only the trivial element of the fundamental group is mapped to e0. Equivalently,we must show that any loop that lifts to an honest loop in S based at e0 is null-homotopic in RP(2), but this is obvious. The lifted loop is homotopic to the constant map in S; choosing a null- homotopy and projecting to RP(2) finishes the proof. These remarks prove that the fundamental group of (RP(2),x0) and the fiber over x0 are equivalent *as sets*; in particular, they have the same cardinality. Since the fiber over any point in our example consists of a pair of antipodal points in S, the fundamental group of (RP(2),x0) is of order 2 and is therefore isomorphic to Z/2Z ( this being the unique group with 2 elements ). In more general cases, this argument clearly doesn't suffice ( e.g., let the cardinality of the fiber be at least 4 ) & one must work a bit harder. The key is that the fiber over any point has a natural group structure arising from the geometry & our function turns out to be an isomorphism of groups. To carry out a description of these ideas in our particular example, we would have to consider the group of all *deck transformations* for the covering map p:S --> RP(2) ; these are maps D:S --> S with the property that pD=p & it is clear that the set of all such maps forms a group with respect to the operation of functional composition. Obviously, if D is one of these maps, then D(e0) lies in the same fiber as does e0; hence, evaluation at e0 defines a function from the group of deck transformations to the fiber over x0. This function is easily proved to be monic -- this is analoguous to the proof that loops lift uniquely. It is also true that this function is onto the fiber & so identifies the fiber over x0 with the group of deck transformations. In view of our previous result, we obtain a set isomorphism between the fundamental group of (RP(2),x0) and the group of deck transformations of the covering map p:S --> RP(2). EXERCISE: This is actually an isomorphism of groups ( the proof of which amounts to showing that the constructions respects the group structures present, which is not at all difficult ). Since, in the present example, the only deck transformations are the identity map and the antipodal map, we have a somewhat more involved verification that the fundamental group is Z/2Z. Ed Hook @ BTL, Piscataway